# The sum of the square of a positive number and the square of 5 more than the number is 277. What...

## Question:

The sum of the square of a positive number and the square of 5 more than the number is 277.

What is the number?

## Quadratic polynomials and Their Solutions:

A quadratic equation is a mathematical equation of degree 2 means the highest power of a variable is one.

General quadratic equation-

{eq}\displaystyle px^{2}+qx+r = 0 {/eq}

here {eq}p \neq 0 {/eq}

here **p**,**r**, and **q** are the real constant

**x** is a variable

A quadratic equation has two solutions (real or complex)

The nature of the quadratic equation is determined by the value of the discriminant (D).

Discriminant {eq}\displaystyle (D) = q^{2}-4pr {/eq}

#### Quadratic Formula

The solutions of the quadratic equation {eq}\displaystyle px^{2}+qx+r = 0 {/eq} are given by the quadratic formula as written below-

{eq}\displaystyle x = \frac{-q \pm \sqrt{q^{2}-4pr}}{2p} {/eq}

## Answer and Explanation:

Let **n** be a number.

Given that the sum of the square of a positive number and the square of 5 more than the number is 277.

{eq}\displaystyle n^{2}+(n+5)^{2} = 277 {/eq}

{eq}\displaystyle n^{2}+n^{2}+10n+25-277 = 0 {/eq}

{eq}\displaystyle 2n^{2}+10n-252 = 0 {/eq}

{eq}\displaystyle n^{2}+5n-126 = 0 {/eq}

using the quadratic formula-

{eq}\displaystyle n = \frac{-5 \pm \sqrt{5^{2}-4(1)(-126)}}{2(1)} {/eq}

{eq}\displaystyle n = \frac{-5 \pm \sqrt{529}}{2} {/eq}

{eq}\displaystyle n = \frac{-5 \pm 23}{2} {/eq}

So

{eq}\displaystyle n = \frac{-5 + 23}{2} = 9 {/eq}

or

{eq}\displaystyle n = \frac{-5 - 23}{2} = -14 {/eq}

Because **n** is a positive number so {eq}n = 9 {/eq} is the right answer.

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#### Learn more about this topic:

from Math 101: College Algebra

Chapter 4 / Lesson 10